Let $R$ be the region enclosed by the line $x=4$, the $x$ -axis, and the curve $y=\sqrt x$. $y$ $x$ ${y=\sqrt{x}}$ $ 4$ $ R$ $ 0$ $ 2$ A solid is generated by rotating $R$ about the line $x=4$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Answer: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\sqrt{x}}$ Notice the slices are horizontal, because we are rotating $R$ about a vertical axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\sqrt{x}}$ $ 4$ $ 0$ $ 2$ $r$ The radius is equal to the distance between the curve ${y=\sqrt x}$ and the line ${x=4}$. To find it, we need to solve $y=\sqrt x$ for $x$ : ${x=y^2}$ So, for any $y$ -value, this is the equation for $r(y)$ : $\begin{aligned} r(y)}=4-y^2} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left(4-y^2}\right)^2 \\\\ &=\pi(16-8y^2+ y^4) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=2$. So the interval of integration is $[0,2]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^2 \pi(16-8y^2+ y^4)\,dy \\\\ &=\pi\int_0^2 (16-8y^2+ y^4)\,dy \end{aligned}$ Let's evaluate the integral. $\pi\int_0^2 (16-8y^2+ y^4)\,dy$ In conclusion, the volume of the solid is $\dfrac{256\pi}{15}$.